In a three level memory hierarchy, the access time of cache, main and virtual memory is 5 nano-seconds, 100 nano-seconds and 10 milli-seconds respectively. If the hit ratio is 80% for the cache and 99.5% for the main memory, then the closest average access time of memory hierarchy in nano-seconds is:

Question

Accepted Answer

The correct option is (3): 10024

The calculation of the average access time ($T_{avg}$) for a multi-level memory hierarchy must account for the access time and the probability of a miss at each level. The general formula for a three-level hierarchy (Cache, Main Memory, and Disk/Virtual Memory) is: $T_{avg} = H_C 	imes T_C + (1 - H_C) 	imes [H_M 	imes T_M + (1 - H_M) 	imes T_{Disk}]$. The first step is to convert all access times into a common unit, nanoseconds (ns): $T_{Disk} = 10 	ext{ ms} = 10 	imes 10^6 	ext{ ns} = 10,000,000 	ext{ ns}$. Substituting the given values into the formula yields: $T_{avg} = (0.8 	imes 5) + (1 - 0.8) 	imes [0.995 	imes 100 + (1 - 0.995) 	imes 10,000,000]$. This calculation simplifies to $T_{avg} = 4 + 0.2 	imes [99.5 + 0.005 	imes 10,000,000]$, which evaluates to $4 + 0.2 	imes (99.5 + 50,000)$, or $4 + 10,019.9 = 10,023.9 	ext{ ns}$. Therefore, the closest integer value is $10024 	ext{ ns}$.

The incorrect option is (1): 512
This result is significantly smaller than the correct value, indicating a major underestimation of the penalty associated with a cache miss that propagates down to the slower memory levels. It suggests a fundamental conceptual error, such as a severe miscalculation of the unit conversion (e.g., mistaking milliseconds for microseconds or an entirely different base) or ignoring the compounding nature of the miss penalties. The calculated time must be dominated by the $10 \text{ ms}$ disk access time, even with a very high hit ratio in the main memory, which is clearly not reflected in this small value.

The incorrect option is (2): 1024
Obtaining a result around 1024 often implies an error in the unit conversion of the disk access time. A student might have incorrectly used a value that is close to a power of 2, perhaps miscalculating the milliseconds to nanoseconds conversion by a large factor, or they may have severely misinterpreted the hit ratios. The final calculated penalty from the main memory and disk levels should be $10,019.9 \text{ ns}$, meaning $1024 \text{ ns}$ is far too small and misses the enormous performance hit caused by the $0.5$ percent of main memory accesses that result in a page fault to the disk.

The incorrect option is (4): 128
This extremely low value is the result of a profound error where the virtual memory/disk access time is completely ignored in the calculation. If a student only accounted for the cache and main memory levels, they would calculate $T\_{avg}$ to be in a much smaller range (e.g., $T\_C + (1-H\_C) T\_M = 5 + 0.2 \times 100 = 25 \text{ ns}$). A final answer of $128 \text{ ns}$ is still too low, but it conceptually shares the same error by neglecting the impact of the $10,000,000 \text{ ns}$ disk access. Even a small miss rate ($0.5$ percent) on the main memory translates to a massive time penalty, which the value $128 \text{ ns}$ fails to incorporate.